$$\int_C \frac{\cos(\ z)}{(z)^2} dz$$
where C is any circle enclosing the origin and oriented counter-clockwise.
z0 = o of order 2 , f(z) = cos z
$$\int_C \frac{\cos(\ z)}{z^2} dz$$ = $2 \pi i f'(0) $ = $2 \pi i (-sin 0) $ = 0
Is this the correct way to approach the problem
Since the Taylor series of $\cos z$ at $0$ is $1 - \dfrac{z^2}{2} + \dfrac{z^4}{24} + \cdots$ the Laurent series of $\dfrac{\cos z}{z^2}$ at $0$ is $\dfrac{1}{z^2} - \dfrac{1}{2} + \dfrac{z^2}{24} + \cdots$ The coefficient of $\dfrac 1z$ in this Laurent series is $0$, so that the residue at $0$ is $0$.